如何更快地迭代十个列表（每个十个元素）的笛卡尔积？ （概率和骰子）

Question

I'm trying to solve this task . I wrote function for this purpose which uses itertools.product() for Cartesian product of input iterables:

def probability(dice_number, sides, target):
    from itertools import product
    from decimal import Decimal
    FOUR_PLACES = Decimal('0.0001')
    total_number_of_experiment_outcomes = sides ** dice_number
    target_hits = 0
    sides_combinations = product(range(1, sides+1), repeat=dice_number)
    for side_combination in sides_combinations:
        if sum(side_combination) == target:
            target_hits += 1
    p = Decimal(str(target_hits / total_number_of_experiment_outcomes)).quantize(FOUR_PLACES)
    return float(p)

When calling probability(2, 6, 3) output is 0.0556 , so works fine. But calling probability(10, 10, 50) calculates veeery long (hours?), but there must be a better way:)
for side_combination in sides_combinations: takes to long to iterate through huge number of sides_combinations .

Please, can you help me to find out how to speed up calculation of result, i want too sleep tonight..

Answer 1

I guess the problem is to find the distribution of the sum of dice. An efficient way to do that is via discrete convolution. The distribution of the sum of variables is the convolution of their probability mass functions (or densities, in the continuous case). Convolution is an n-ary operator, so you can compute it conveniently just two pmf's at a time (the current distribution of the total so far, and the next one in the list). Then from the final result, you can read off the probabilities for each possible total. The first element in the result is the probability of the smallest possible total, and the last element is the probability of the largest possible total. In between you can figure out which one corresponds to the particular sum you're looking for.

The hard part of this is the convolution, so work on that first. It's just a simple summation, but it's just a little tricky to get the limits of the summation correct. My advice is to work with integers or rationals so you can do exact arithmetic.

After that you just need to construct an appropriate pmf for each input die. The input is just [1, 1, 1, ... 1] if you're using integers (you'll have to normalize eventually) or [1/n, 1/n, 1/n, ..., 1/n] if rationals, where n = number of faces. Also you'll need to label the indices of the output correctly -- again this is just a little tricky to get it right.

Convolution is a very general approach for summations of variables. It can be made even more efficient by implementing convolution via the fast Fourier transform, since FFT(conv(A, B)) = FFT(A) FFT(B). But at this point I don't think you need to worry about that.

Answer 2

If someone still interested in solution which avoids very-very-very long iteration process through all itertools.product Cartesian products, here it is:

def probability(dice_number, sides, target):
    if dice_number == 1:
        return (1 <= target <= sides**dice_number) / sides
    return sum([probability(dice_number-1, sides, target-x) \
                        for x in range(1,sides+1)]) / sides

But you should add caching of probability function results, if you won't - calculation of probability will takes very-very-very long time as well)

PS this code is 100% not mine, i took it from the internet, i'm not such smart to product it by myself, hope you'll enjoy it as much as i.